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kodGreya [7K]
3 years ago
13

A raindrop has a mass of 50. mg and the Pacific Ocean has a mass of 7.08 x 10^20 kg. what is the mass of 1 mole of raindrops? Ho

w many moles of raindrops are in the Pacific Ocean?
Chemistry
2 answers:
Anna007 [38]3 years ago
7 0

Solution:

According to the Avogadro's number:  

6.022 *10^23 drops per 0.050 g/ drop = 3.011 *10^22 grams per mole of drops  

3.011 *10^22 grams per 1 kg / 1000 grams = 3.011 *10^19 kilograms / mole of drops  

thus the answers are:  

3.0 *10^22 grams per mole of drops  

3.0 *10^19 kilograms per mole of drops  

And,

In the calculation of how many moles of raindrops in the Pacific Ocean is:  

7.08X10^20kg per 3.0 *10^19 kilograms per mole of drops = 23.5 moles of drops

This is the required solution.  


Ahat [919]3 years ago
7 0

Raindrops vary in size from about 0.02 in. (0.5 mm) to as much as 0.33 in. (8 mm) in thunderstorms. Raindrops come in many different sizes and can get to as big as a quarter inch. They cannot get bigger than that and the reason is that the wind resistance begins breaking them up into smaller raindrops. Raindrops don't get much larger than a quarter inch

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2 years ago
The solubility of silver(I)phosphate at a given temperature is 1.02 g/L. Calculate the Ksp at this temperature. After you get yo
Snezhnost [94]

<u>Answer:</u> The solubility product of silver (I) phosphate is 9.57\times 10^{-10}

<u>Explanation:</u>

We are given:

Solubility of silver (I) phosphate = 1.02 g/L

To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:

Molar mass of silver (I) phosphate = 418.6 g/mol

\text{Molar solubility of silver (I) phosphate}=\frac{1.02g/L}{418.6g/mol}=2.44\times 10^{-3}mol/L

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The chemical equation for the ionization of silver (I) phosphate follows:

Ag_3PO_4(aq.)\rightleftharpoons 3Ag^{+}(aq.)+PO_4^{3-}(aq.)  

                            3s                  s

The expression of K_{sp} for above equation follows:

K_{sp}=(3s)^3\times s

We are given:  

s=2.44\times 10^{-3}M

Putting values in above expression, we get:

K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}

Hence, the solubility product of silver (I) phosphate is 9.57\times 10^{-10}

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The rate constant for a particular zero-order reaction is 0.075 M s-1. If the initial concentration of reactant is 0.537 M it ta
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Answer:

It takes 5.83s to decrease the concentration of the reactant from 0.537M to 0.100M

Explanation:

A zero-order reaction follows the equation:

[A] = [A]₀ - kt

<em>Where [A] is actual reaction of the reactant = 0.100M</em>

<em>[A]₀ the initial concentration = 0.537M</em>

<em>k is rate constant = 0.075Ms⁻¹</em>

<em>And t is time it takes:</em>

<em />

0.100M = 0.537M -0.075Ms⁻¹t

-0.437M = -0.075Ms⁻¹t

5.83s = t

It takes 5.83s to decrease the concentration of the reactant from 0.537M to 0.100M

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